Home/ Intrebari/Q 92807
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

Penguin200
user (0)
Pe: 2018-07-25T07:45:32+03:00 In: In:

Limite sume

Buna!Oare cum as putea rezolva urmatoarea problema cu cunostiinte de nivelul clasei a 11-a?
Cunoscand \lim_{x \to inf}\sum_{k=1}^{n} \frac{1}{k^2}=\frac{pi^2}{6}, care este valoarea \lim_{x\to inf}\sum_{k=1}^{n} \frac{1}{(2k-1)^2}

Multumesc!

Similare

1 raspuns

  1. PhantomR
    expert (VI)

    Salut,

    Daca notam sirul s_n=\sum_{k=1}^n \frac{1}{k^2}, avem ca \lim s_n = \frac{\pi^2}{6}.

    Limita de calculat se scrie \lim \sum_{k=1}^n \frac{1}{(2k-1)^2}=\lim \frac{1}{1^2}+\frac{1}{3^2}+...+\frac{1}{(2n-1)^2}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{(2n-1)^2}-(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{(2n-2)^2})=
    =\sum_{k=1}^{2n-1} \frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{(2k)^2}=\sum_{k=1}^{2n-1} \frac{1}{k^2}-\frac{1}{2^2}\sum_{k=1}^{n-1}\frac{1}{k^2}=s_{2n-1}-\frac{1}{4}s_{n-1}.

    Trecand la limita si folosind faptul ca subsirurile (s_{2n-1}),(s_{n-1}) au aceeasi limita ca si sirul (s_n), obtinem ca limita cautata este \frac{3}{4} \frac{\pi^2}{6}=\frac{\pi^2}{4}

    • 0
Raspunde

Raspunde